Tickets are numbered from 1 to 18 are mixed up together and then 9 tickets are drawn at random. Find the probability that the ticket has a number, which is a multiple of 2 or 3.

S = { 1, 2, 3, 4, .....18 } 

=> n(S) = 18

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

  $E3 =E1\cap E2=\{6, 12, 18\}$ 

=> n(E3) = 3

$\therefore E=E1 \cup E2 = E1+E2-E3$ 

=> n(E) = 9 + 6 - 3 =12

 where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{12}{18}=\frac{2}{3}$